Sling Tension on a Two-Leg Bridle at a Tight Angle
A precast concrete panel weighing 3.2 t is to be lifted by two chain slings in a bridle from a single hook. Each leg attaches to a cast-in anchor at the top corners. The horizontal distance between the anchors is 2.4 m, and the vertical distance from the hook to the anchor line is 1.0 m. The site supervisor says the angle is 'fine, it's only a bit wide' and wants to proceed. The chains are rated 8 t WLL in straight lift. The ground is level, the panel is free of obstructions, and the exclusion zone is marked.
1. Calculate the angle each leg makes with the vertical and the tension in each leg. State whether the 8 t chains are adequate on a straight-lift basis.
Read the worked answer
Each leg runs from the hook to an anchor. Half the horizontal span is 1.2 m, vertical drop is 1.0 m. The angle from vertical θ satisfies tan θ = 1.2/1.0 = 1.2, so θ ≈ 50.2°. The leg length is √(1.2² + 1.0²) = √(1.44 + 1.0) = √2.44 ≈ 1.56 m.
For a symmetric two-leg bridle, each leg carries half the vertical load divided by cos θ. Half the load is 1.6 t. Tension per leg = 1.6 / cos 50.2° = 1.6 / 0.640 ≈ 2.5 t.
So each 8 t chain is carrying about 2.5 t, well within its straight-lift rating. But the angle is 50° from vertical, which is a 100° included angle between legs - beyond the common 90° guidance and into the range where many chains must be derated. The correct check is against the manufacturer's angle derating: at 100° included, the multiplier is roughly 1/cos 50° ≈ 1.56, so the effective capacity is about 8/1.56 ≈ 5.1 t per leg, still adequate here. The supervisor's 'it's fine' is not a method; the angle must be measured and the derating applied.
2. The panel is later found to have its CG 0.3 m off centre toward one anchor. Recalculate the leg tensions and explain the consequence.
Read the worked answer
With the CG offset toward one anchor, the vertical reactions are no longer equal. Taking moments about the far anchor over a 2.4 m span: the near anchor carries (1.2 + 0.3)/2.4 of the load = 1.5/2.4 = 0.625, and the far anchor carries 0.375. So near vertical reaction = 0.625 × 3.2 = 2.0 t; far = 1.2 t.
Each leg tension is the vertical reaction divided by cos θ, with θ still about 50.2° and cos θ ≈ 0.640. Near leg = 2.0/0.640 ≈ 3.1 t; far leg = 1.2/0.640 ≈ 1.9 t.
The near chain now carries about 3.1 t, a 24% increase over the symmetric case. Against the derated capacity of about 5.1 t it is still acceptable, but the margin has narrowed and the panel will hang tilted. The lesson is that a CG offset multiplies with the sling angle: both effects compound on the same leg. The panel should be re-rigged so the hook is over the CG, or the offset accounted for in the rigging plan, before the lift proceeds.